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1. • CommentRowNumber1.
   • CommentAuthorDavidRoberts
   • CommentTimeMay 29th 2012
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexGiven a category of smooth spaces, containing a full subcategory of Frechet manifolds and a compatible 'underlying topological space' functor, what is the definition of a regular map, and how does it relate to a regular map on the subcategory of Frechet manifolds? I'm more interested in submersions, so how would one define a submersion in a category of smooth spaces? One way to do it in diffeological spaces is to consider subductions (maps which are submersions when restricted to each chart), but I don't think these give anything nice for Frechet manifolds. Another thing, is there a reasonable definition of a pair of transversal maps of smooth spaces with common codomain? (PS: [this MO question](http://mathoverflow.net/questions/98160/regular-maps-between-frechet-manifolds-and-pullbacks))

   Given a category of smooth spaces, containing a full subcategory of Frechet manifolds and a compatible ’underlying topological space’ functor, what is the definition of a regular map, and how does it relate to a regular map on the subcategory of Frechet manifolds? I’m more interested in submersions, so how would one define a submersion in a category of smooth spaces? One way to do it in diffeological spaces is to consider subductions (maps which are submersions when restricted to each chart), but I don’t think these give anything nice for Frechet manifolds.

   Another thing, is there a reasonable definition of a pair of transversal maps of smooth spaces with common codomain?

   (PS: this MO question)

2. • CommentRowNumber2.
   • CommentAuthorAndrew Stacey
   • CommentTimeMay 29th 2012
   • PermaLink
   Author: Andrew Stacey
   Format: MarkdownItexThe honest answer is: I don't know! It depends what result you are trying to prove. It would seem that you want to conclude that certain things are smooth Frechet manifolds. In that case, ultimately you want to conclude stuff about charts. In that case, you have two avenues open to you: direct construction or inverse function theorem. However, the IVT looks quite horrible for Frechet spaces! Regarding the definition of a regular map in a category of generalised smooth spaces, I don't think it makes much sense there. Regular maps in (finite dimensional) manifolds are so intertwined with charts that I don't know what properties I'd want for generalised smooth spaces. Subduction looks nice (and can be generalised to any "maps in") but I'd worry that you might need more than that: I don't see (haven't thought *too* much about it) why the lifts would automatically be smooth. And that's something that one would get with a "proper" regular map. So I think the best strategy is to work with the maps you have and see what happens in a case-by-case analysis. Regarding your (email) question: > given a regular map of smocally $\mathcal{T}$-compact smooth spaces $p \colon C \to D$, what is happening with the map $X^D \to X^C$ given by precomposition? The word "regular" doesn't make sense here, by what I've just said. The set-up means that you'll get nice charts on $X^D$ and $X^C$ so that the map $X^D \to X^C$ ends up being a linear map. What you'd then like is for that to split. Having a section $D \to C$ ought to be enough for that. Then probably having local sections would be enough.

   The honest answer is: I don’t know! It depends what result you are trying to prove. It would seem that you want to conclude that certain things are smooth Frechet manifolds. In that case, ultimately you want to conclude stuff about charts. In that case, you have two avenues open to you: direct construction or inverse function theorem. However, the IVT looks quite horrible for Frechet spaces!

   Regarding the definition of a regular map in a category of generalised smooth spaces, I don’t think it makes much sense there. Regular maps in (finite dimensional) manifolds are so intertwined with charts that I don’t know what properties I’d want for generalised smooth spaces. Subduction looks nice (and can be generalised to any “maps in”) but I’d worry that you might need more than that: I don’t see (haven’t thought too much about it) why the lifts would automatically be smooth. And that’s something that one would get with a “proper” regular map.

   So I think the best strategy is to work with the maps you have and see what happens in a case-by-case analysis.

   Regarding your (email) question:

   given a regular map of smocally $\mathcal{T}$-compact smooth spaces $p \colon C \to D$, what is happening with the map $X^D \to X^C$ given by precomposition?

   The word “regular” doesn’t make sense here, by what I’ve just said. The set-up means that you’ll get nice charts on $X^D$ and $X^C$ so that the map $X^D \to X^C$ ends up being a linear map. What you’d then like is for that to split. Having a section $D \to C$ ought to be enough for that. Then probably having local sections would be enough.

3. • CommentRowNumber3.
   • CommentAuthorAndrew Stacey
   • CommentTimeMay 29th 2012
   • PermaLink
   Author: Andrew Stacey
   Format: MarkdownItexTo turn to Raymond's email, the whole point of that preprint (now available [here](http://www.math.ntnu.no/~stacey/Research/Preprints/smocally.html)) was to define a simple context in which taking mapping spaces "just worked". So "smoothly regular" implies "regular" by construction, just as "locally additive" implies "manifold". The same holds for submersion and immersion. The hard part is showing that for finite dimensional manifolds, "regular" in the usual sense implies "smoothly regular". In brief, * Regular $\implies$ "nice" charts at every point * Smoothly regular $\implies$ "nice" charts at every point *simultaneously* Immersion and submersion will follow, since each is a regular map where the "nice" charts are especially nice, but that especial niceness is actually independent of the chart: if I can find charts such that my submersion looks like a linear projection, then that linear projection *must* be full rank.

   To turn to Raymond’s email, the whole point of that preprint (now available here) was to define a simple context in which taking mapping spaces “just worked”. So “smoothly regular” implies “regular” by construction, just as “locally additive” implies “manifold”. The same holds for submersion and immersion. The hard part is showing that for finite dimensional manifolds, “regular” in the usual sense implies “smoothly regular”.

   In brief,

   • Regular $\implies$ “nice” charts at every point
   • Smoothly regular $\implies$ “nice” charts at every point simultaneously

   Immersion and submersion will follow, since each is a regular map where the “nice” charts are especially nice, but that especial niceness is actually independent of the chart: if I can find charts such that my submersion looks like a linear projection, then that linear projection must be full rank.

4. • CommentRowNumber4.
   • CommentAuthorDavidRoberts
   • CommentTimeMay 29th 2012
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   Author: DavidRoberts
   Format: MarkdownItexThanks, Andrew. I guess I was under the impression that you were considering regular maps of smooth spaces, and I was concerned that I hadn't seen a definition. I'm quite pedestrian in my approach to differential geometry, so sometimes I need to see there things spelled out in words of one syllable (much like the famous proof of Lob's theorem:-)

   Thanks, Andrew. I guess I was under the impression that you were considering regular maps of smooth spaces, and I was concerned that I hadn’t seen a definition. I’m quite pedestrian in my approach to differential geometry, so sometimes I need to see there things spelled out in words of one syllable (much like the famous proof of Lob’s theorem:-)

5. • CommentRowNumber5.
   • CommentAuthorDavidRoberts
   • CommentTimeMay 29th 2012
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexCrossed comments there. I think I may also be getting confused with the overloading of the term 'regular', since it applies to spaces and maps, in different ways. Well, to bed, and thinking about it some more tomorrow.

   Crossed comments there. I think I may also be getting confused with the overloading of the term ’regular’, since it applies to spaces and maps, in different ways. Well, to bed, and thinking about it some more tomorrow.

6. • CommentRowNumber6.
   • CommentAuthorAndrew Stacey
   • CommentTimeMay 29th 2012
   • PermaLink
   Author: Andrew Stacey
   Format: MarkdownItexTo clarify: I'm only ever using "regular" (or its variants) for maps. So in my summary above, that was "nice charts wrt the function" in both cases. But to talk about "nice charts wrt the function" you first need a notion of "nice charts". So the full list is: * manifold $\implies$ linear charts at each point * locally additive $\implies$ linear charts at each point simultaneously * regular map $\implies$ linear charts at each point st map is linear projection * smoothly regular map $\implies$ linear charts at each point simultaneously st map is linear projection

   To clarify: I’m only ever using “regular” (or its variants) for maps. So in my summary above, that was “nice charts wrt the function” in both cases. But to talk about “nice charts wrt the function” you first need a notion of “nice charts”. So the full list is:

   • manifold $\implies$ linear charts at each point
   • locally additive $\implies$ linear charts at each point simultaneously
   • regular map $\implies$ linear charts at each point st map is linear projection
   • smoothly regular map $\implies$ linear charts at each point simultaneously st map is linear projection
7. • CommentRowNumber7.
   • CommentAuthorDavidRoberts
   • CommentTimeMay 30th 2012
   • (edited May 30th 2012)
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexYour links, Andrew, on the page for 'smocally.pdf' do not all point to that file! The lower links are to the paper on piecewise smooth paths. Re: 'regular' - ah, that clarifies things!

   Your links, Andrew, on the page for ’smocally.pdf’ do not all point to that file! The lower links are to the paper on piecewise smooth paths.

   Re: ’regular’ - ah, that clarifies things!

8. • CommentRowNumber8.
   • CommentAuthorAndrew Stacey
   • CommentTimeMay 30th 2012
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   Author: Andrew Stacey
   Format: MarkdownItexWhoops! That's what comes of cut-and-paste. Fixed now (I hope).

   Whoops! That’s what comes of cut-and-paste. Fixed now (I hope).

9. • CommentRowNumber9.
   • CommentAuthorDavidRoberts
   • CommentTimeMay 31st 2012
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexIf we revisit this question > given a regular map of smocally $\mathcal{T}$-compact smooth spaces $p \colon C \to D$, what is happening with the map $X^D \to X^C$ given by precomposition? where instead of a map of smooth spaces $C\to D$ we have a submersion between finite dimensional, compact smooth manifolds, can we say anything? Clearly $X^D \to X^C$ is an inclusion, but I have no feeling for what it looks like. I suspect it is a closed submanifold.

   If we revisit this question

   given a regular map of smocally $\mathcal{T}$-compact smooth spaces $p \colon C \to D$, what is happening with the map $X^D \to X^C$ given by precomposition?

   where instead of a map of smooth spaces $C\to D$ we have a submersion between finite dimensional, compact smooth manifolds, can we say anything? Clearly $X^D \to X^C$ is an inclusion, but I have no feeling for what it looks like. I suspect it is a closed submanifold.

10. • CommentRowNumber10.
    • CommentAuthorAndrew Stacey
    • CommentTimeMay 31st 2012
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexRight, so let's fix our setup. Let $X$ be a smoothly locally additive space. So that means that we have a neighbourhood $V \subseteq X \times X$ of the diagonal with smooth maps $\mathbb{R} \times V \to V$ and $V \times_X V \to V$ making the fibres of $\pi_1 \colon V \to X$ into vector spaces. Then we have smoothly $\mathcal{T}$-compact spaces $C$ and $D$ with a map $p \colon C \to D$. The inheritance result says that $X^C$ and $X^D$ (mapping objects in our category of smooth spaces) are again smoothly locally additive, with neighbourhoods $V^C$ and $V^D$. The map $p \colon C \to D$ defines a map $p^* \colon X^D \to X^C$ by composition. There is an obvious diagram: $$ \begin{matrix} V^{\mathrlap{D}} & \to & V^{\mathrlap{C}} \\ \downarrow & & \downarrow \\ X^D \times X^D & \to & X^C \times X^C \end{matrix} $$ So let's take an element $\alpha \colon D \to X$. Then the smoothly locally additive structure gives us a chart for $X^D$ at $\alpha$ given by $\{\beta \colon D \to V : \pi_1 \beta = \alpha\} = \Gamma_D(\alpha^* V)$. This is a smooth vector space with pointwise operations. Now $p^* \alpha = \alpha \circ p$ is a smooth map $C \to X$ and we get a chart for $X^C$ at $p^* \alpha$ with domain $\{\gamma \colon C \to V : \pi_1 \gamma = p^* \alpha\}$. In these charts, the map $p^*$ is given again by composition: $(\beta \colon D \to V) \mapsto (\beta \circ p \colon C \to V)$. This is a linear map. This means that $p^* \colon X^D \to X^C$ is a map of smoothly locally additive spaces and so is what I called "smoothly regular". Now, in finite dimensions, any linear map $T \colon \mathbb{R}^n \to \mathbb{R}^m$ can be regarded as a projection in that we can choose bases for the source and target so that $T$ is $(x_1,\dots,x_n) \mapsto (x_1,\dots,x_k,0,\dots,0)$. But in so doing we are implicitly regarding $\mathbb{R}^k \subseteq \mathbb{R}^n$ and $\mathbb{R}^k \subseteq \mathbb{R}^m$ as the same space. Removing ourselves from the dependency of bases, what we're trying to say is that whenever we have a linear map $T \colon E \to F$ then we have a (not exact) sequence: $$ \ker T \hookrightarrow E \twoheadrightarrow E/\ker T \to \im T \hookrightarrow F \to F/\im T $$ In the best of all possible worlds, each map splits. This means that $E \cong \ker T \oplus E/\ker T$, $E/\ker T \cong \im T$, and $F \cong \im T \oplus F/\im T$. This is automatic in finite dimensions. For Banach spaces, we need to know that $T$ is closed. But we're dealing with smooth vector spaces so life is more complicated. To get into this "ideal situation", the simplest route is to construct a smooth linear map $S \colon F \to E$ with the property that $T S T = T$ and $S T S = S$. In this situation, $S T$ is a projection onto $\im T$ and $I - T S$ is a projection onto $\ker T$. Then everything splits as desired. In our case, we need $S \colon V^C \to V^D$ with the property that $S(\beta) \circ p = \beta$. There are several ways to construct such a map, depending on what we know about our source spaces. I think that if $p \colon C \to D$ is a regular map of compact manifolds then one can construct a splitting map using some partition-of-unity argument - I'll give this some thought. Assuming that $S$ exists, then we have the best of all possible worlds. If $\ker T = \{0\}$ then $T$ is an embedding onto a closed subspace, whereupon in the manifold world we have a closed submanifold. If $\im T = F$ then $T$ is a projection onto a closed subspace, whereupon in the manifold world we have the best kind of submersion. So it hinges on knowing when $V^D \to V^C$ has what I really want to call an adjoint.

    Right, so let’s fix our setup. Let $X$ be a smoothly locally additive space. So that means that we have a neighbourhood $V \subseteq X \times X$ of the diagonal with smooth maps $\mathbb{R} \times V \to V$ and $V \times_X V \to V$ making the fibres of $\pi_1 \colon V \to X$ into vector spaces. Then we have smoothly $\mathcal{T}$-compact spaces $C$ and $D$ with a map $p \colon C \to D$. The inheritance result says that $X^C$ and $X^D$ (mapping objects in our category of smooth spaces) are again smoothly locally additive, with neighbourhoods $V^C$ and $V^D$.

    The map $p \colon C \to D$ defines a map $p^* \colon X^D \to X^C$ by composition. There is an obvious diagram:

    $$\begin{matrix} V^{\mathrlap{D}} & \to & V^{\mathrlap{C}} \\ \downarrow & & \downarrow \\ X^D \times X^D & \to & X^C \times X^C \end{matrix}$$

    So let’s take an element $\alpha \colon D \to X$. Then the smoothly locally additive structure gives us a chart for $X^D$ at $\alpha$ given by $\{\beta \colon D \to V : \pi_1 \beta = \alpha\} = \Gamma_D(\alpha^* V)$. This is a smooth vector space with pointwise operations. Now $p^* \alpha = \alpha \circ p$ is a smooth map $C \to X$ and we get a chart for $X^C$ at $p^* \alpha$ with domain $\{\gamma \colon C \to V : \pi_1 \gamma = p^* \alpha\}$. In these charts, the map $p^*$ is given again by composition: $(\beta \colon D \to V) \mapsto (\beta \circ p \colon C \to V)$. This is a linear map. This means that $p^* \colon X^D \to X^C$ is a map of smoothly locally additive spaces and so is what I called “smoothly regular”.

    Now, in finite dimensions, any linear map $T \colon \mathbb{R}^n \to \mathbb{R}^m$ can be regarded as a projection in that we can choose bases for the source and target so that $T$ is $(x_1,\dots,x_n) \mapsto (x_1,\dots,x_k,0,\dots,0)$. But in so doing we are implicitly regarding $\mathbb{R}^k \subseteq \mathbb{R}^n$ and $\mathbb{R}^k \subseteq \mathbb{R}^m$ as the same space. Removing ourselves from the dependency of bases, what we’re trying to say is that whenever we have a linear map $T \colon E \to F$ then we have a (not exact) sequence:

    $$\ker T \hookrightarrow E \twoheadrightarrow E/\ker T \to \im T \hookrightarrow F \to F/\im T$$

    In the best of all possible worlds, each map splits. This means that $E \cong \ker T \oplus E/\ker T$, $E/\ker T \cong \im T$, and $F \cong \im T \oplus F/\im T$. This is automatic in finite dimensions. For Banach spaces, we need to know that $T$ is closed. But we’re dealing with smooth vector spaces so life is more complicated. To get into this “ideal situation”, the simplest route is to construct a smooth linear map $S \colon F \to E$ with the property that $T S T = T$ and $S T S = S$. In this situation, $S T$ is a projection onto $\im T$ and $I - T S$ is a projection onto $\ker T$. Then everything splits as desired.

    In our case, we need $S \colon V^C \to V^D$ with the property that $S(\beta) \circ p = \beta$. There are several ways to construct such a map, depending on what we know about our source spaces. I think that if $p \colon C \to D$ is a regular map of compact manifolds then one can construct a splitting map using some partition-of-unity argument - I’ll give this some thought.

    Assuming that $S$ exists, then we have the best of all possible worlds. If $\ker T = \{0\}$ then $T$ is an embedding onto a closed subspace, whereupon in the manifold world we have a closed submanifold. If $\im T = F$ then $T$ is a projection onto a closed subspace, whereupon in the manifold world we have the best kind of submersion.

    So it hinges on knowing when $V^D \to V^C$ has what I really want to call an adjoint.

11. • CommentRowNumber11.
    • CommentAuthorDavidRoberts
    • CommentTimeMay 31st 2012
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexOf course, in the case we are interested in ($p$ a surjective submersion), there are local sections through every point of $C$, so I can sort of see how partitions of unity might come in handy.

    Of course, in the case we are interested in ($p$ a surjective submersion), there are local sections through every point of $C$, so I can sort of see how partitions of unity might come in handy.

12. • CommentRowNumber12.
    • CommentAuthorAndrew Stacey
    • CommentTimeMay 31st 2012
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexRight, so let's imagine we're in a great situation: $C$ and $D$ are closed manifolds, $p \colon C \to D$ is at least regular, and we might go for submersion if we feel like it later. $X$ and $V$ are as above. So we start with $\alpha \colon D \to X$ and $p^* \alpha = \alpha \circ p \colon C \to X$. We get smooth vector spaces $E_D \coloneqq \Gamma_D(\alpha^* V)$ and $E_C \coloneqq \Gamma_C(p^*\alpha^* V)$ and a linear map $E_D \to E_C$ given by $\beta \mapsto \beta \circ p$. We want a map $S \colon E_C \to E_D$. So given a map $\gamma \colon C \to V$ such that $\pi_1 \gamma = \alpha \circ p$ we want $S(\gamma) \colon D \to V$ with $\pi_1 S(\gamma) = \alpha$ and certain nice other relationships. Actually, let's start with a submersion. I think if $p \colon C \to D$ is not surjective then I need some extra assumptions on $V$ (which might be automatically satisfied, but I don't want to think about that yet). So let $p \colon C \to D$ be a surjective submersion. As $C$ and $D$ are closed, this makes $C$ a fibre bundle over $D$. Choose a locally finite trivialisation $\mathcal{U}$ and a subordinate partition of unity, $\{\rho\}$. For $U \in \mathcal{U}$, choose a section $s_U \colon U \to C_U = p^{-1}(U)$. Now for $\gamma \colon C \to V$ with $\pi_1(\gamma) = \alpha \circ p$, we define $S_U(\gamma) \colon U \to V$ by $c \mapsto \rho_U(c) \gamma \circ s_U(c)$. The multiplication is taking place in $V$. Then define $S(\gamma) = \sum_U S_U(\gamma)$. Again, the addition is taking place fibrewise in $V$. This will be smooth for the usual arguments (note: it'll be an assumption on the category of smooth objects that these "usual arguments" still hold). This should do the trick. As the choices all depend on $C$ and $D$, this is globally defined (on $X^C$ and $X^D$) and so gives a smooth choice of splitting map. With a bit more polish, this should show that $X^D$ is a closed submanifold of $X^C$, possibly with a tubular neighbourhood.

    Right, so let’s imagine we’re in a great situation: $C$ and $D$ are closed manifolds, $p \colon C \to D$ is at least regular, and we might go for submersion if we feel like it later. $X$ and $V$ are as above.

    So we start with $\alpha \colon D \to X$ and $p^* \alpha = \alpha \circ p \colon C \to X$. We get smooth vector spaces $E_D \coloneqq \Gamma_D(\alpha^* V)$ and $E_C \coloneqq \Gamma_C(p^*\alpha^* V)$ and a linear map $E_D \to E_C$ given by $\beta \mapsto \beta \circ p$. We want a map $S \colon E_C \to E_D$. So given a map $\gamma \colon C \to V$ such that $\pi_1 \gamma = \alpha \circ p$ we want $S(\gamma) \colon D \to V$ with $\pi_1 S(\gamma) = \alpha$ and certain nice other relationships.

    Actually, let’s start with a submersion. I think if $p \colon C \to D$ is not surjective then I need some extra assumptions on $V$ (which might be automatically satisfied, but I don’t want to think about that yet).

    So let $p \colon C \to D$ be a surjective submersion. As $C$ and $D$ are closed, this makes $C$ a fibre bundle over $D$. Choose a locally finite trivialisation $\mathcal{U}$ and a subordinate partition of unity, $\{\rho\}$. For $U \in \mathcal{U}$, choose a section $s_U \colon U \to C_U = p^{-1}(U)$. Now for $\gamma \colon C \to V$ with $\pi_1(\gamma) = \alpha \circ p$, we define $S_U(\gamma) \colon U \to V$ by $c \mapsto \rho_U(c) \gamma \circ s_U(c)$. The multiplication is taking place in $V$. Then define $S(\gamma) = \sum_U S_U(\gamma)$. Again, the addition is taking place fibrewise in $V$. This will be smooth for the usual arguments (note: it’ll be an assumption on the category of smooth objects that these “usual arguments” still hold).

    This should do the trick. As the choices all depend on $C$ and $D$, this is globally defined (on $X^C$ and $X^D$) and so gives a smooth choice of splitting map.

    With a bit more polish, this should show that $X^D$ is a closed submanifold of $X^C$, possibly with a tubular neighbourhood.

13. • CommentRowNumber13.
    • CommentAuthorDavidRoberts
    • CommentTimeJun 1st 2012
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexHere's another case Ray and I were thinking of, which doesn't quite fit into the general pattern of the discussion so far (and would probably be enough for the first cases of interest). Consider the restriction map $X^{[0,1]} \to X^{[0,1/2]}$. I conjecture this is a submersion, and when $X$ is a manifold without boundary, I make the more tentative conjecture that it is surjective (or, vice versa, that this map would only fail to be surjective if $X$ was a manifold with boundary). I think this should follow from standard results about flows. Since we can write the restriction map as the composition of a series of restriction maps, I believe it should be enough, given a function $f\colon [0,1] \to X$ to consider the case when this lands inside a single chart, namely trying to show that $(\mathbb{R}^n)^{[0,1]} \to (\mathbb{R}^n)^{[0,1/2]}$ is a submersion.

    Here’s another case Ray and I were thinking of, which doesn’t quite fit into the general pattern of the discussion so far (and would probably be enough for the first cases of interest). Consider the restriction map $X^{[0,1]} \to X^{[0,1/2]}$. I conjecture this is a submersion, and when $X$ is a manifold without boundary, I make the more tentative conjecture that it is surjective (or, vice versa, that this map would only fail to be surjective if $X$ was a manifold with boundary). I think this should follow from standard results about flows.

    Since we can write the restriction map as the composition of a series of restriction maps, I believe it should be enough, given a function $f\colon [0,1] \to X$ to consider the case when this lands inside a single chart, namely trying to show that $(\mathbb{R}^n)^{[0,1]} \to (\mathbb{R}^n)^{[0,1/2]}$ is a submersion.

14. • CommentRowNumber14.
    • CommentAuthorAndrew Stacey
    • CommentTimeJun 1st 2012
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexThis is Seeley's result: <http://www.ams.org/mathscinet-getitem?mr=165392> The result is that the space $\{f \colon [0,\infty) \to \mathbb{R}^n\}$ is a direct summand of $C^\infty(\mathbb{R},\mathbb{R}^n)$ from which the result you need can be deduced.

    This is Seeley’s result: http://www.ams.org/mathscinet-getitem?mr=165392

    The result is that the space $\{f \colon [0,\infty) \to \mathbb{R}^n\}$ is a direct summand of $C^\infty(\mathbb{R},\mathbb{R}^n)$ from which the result you need can be deduced.

15. • CommentRowNumber15.
    • CommentAuthorDavidRoberts
    • CommentTimeJun 1st 2012
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    Author: DavidRoberts
    Format: MarkdownItexAwesome! Thanks for that reference.

    Awesome! Thanks for that reference.

16. • CommentRowNumber16.
    • CommentAuthorAndrew Stacey
    • CommentTimeJun 4th 2012
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexHmm, base change will always be a morphism of smoothly locally $P$-spaces so I think "smoothly regular" should be reserved for the situation where the linear map splits - this is necessary for tubular neighbourhoods, for example.

    Hmm, base change will always be a morphism of smoothly locally $P$-spaces so I think “smoothly regular” should be reserved for the situation where the linear map splits - this is necessary for tubular neighbourhoods, for example.